# set3 - Lecture 7 Series Solutions(cont’d Let us now...

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Unformatted text preview: Lecture 7 Series Solutions (cont’d) Let us now consider the so-called Legendre equation, (1 − x 2 ) y ′′ − 2 xy ′ + p ( p + 1) y = 0 , p ∈ R , (1) which occurs quite frequently in applied mathematics and theoretical physics. (In the solu- tion of the Laplace and Schr¨odinger equations in three dimensions, x above represents the angular coordinate θ . In quantum mechanics, the constant p will represent the orbital angular momentum quantum number.) Rewriting this DE in standard form, y ′′ − 2 x 1 − x 2 y ′ + p ( p + 1) 1 − x 2 y = 0 , (2) we have P ( x ) = − 2 x 1 − x 2 , Q ( x ) = p ( p + 1) 1 − x 2 . (3) The functions P ( x ) and Q ( x ) are seen to be analytic at x = 0. The “bad” points of this DE are x = ± 1. As a consequence, the power series of P ( x ) and Q ( x ) have power series expansions about 0 with radius of convergence R = 1. For example, P ( x ) = 2 x 1 − x 2 = 2 x [1 + x 2 + x 4 + ··· ] , | x | < 1 . (4) From the theorem cited in the previous lecture, we expect that power series solutions of (1) having the form y ( x ) = summationdisplay n a n x n (5) will also have radii of convergence R = 1. Let us now work out these series solutions. We substitute the above expansion into the DE in (1) (not paying attention to the lower limits of summation, with the understanding that a n = 0 for n < 0): summationdisplay n ( n − 1) a n x n − 2 − x 2 summationdisplay n ( n − 1) a n x n − 2 − 2 x summationdisplay na n x n − 1 + p ( p + 1) summationdisplay a n x n = 0 . (6) 1 Collecting like terms in x n , and bumping up or down the indices of each summation as necessary: summationdisplay [( n + 2)( n + 1) a n +2 − n ( n − 1) a n − 2 na n + p ( p + 1) a n ] x n = 0 , (7) which can be simplified further to summationdisplay [( n + 2)( n + 1) a n +2 + [ − n ( n + 1) + p ( p + 1)] a n ] x n = 0 . (8) This implies that ( n + 2)( n + 1) a n +2 + [ − n ( n + 1) + p ( p + 1)] a n = 0 (9) for all n . Setting n = 0 gives 2 a 2 + p ( p + 1) a = 0 , or a 2 = − 1 2 p ( p + 1) a . (10) Setting n = 1 gives 6 a 3 + [ − 2 + p ( p + 1) a 1 = 0 , or a 3 = − 1 6 ( p − 1)( p + 2) a 1 . (11) For general n ≥ 0, a n +2 = n ( n + 1) − p ( p + 1) ( n + 1)( n + 2) a n (12) = − ( p − n )( p + n + 1) ( n + 1)( n + 2) a n Let’s compute the next two terms: For n = 2 in (12), we have a 4 = − ( p − 2)( p + 3) 4 · 3 a 2 = p ( p − 2)( p + 1)( p + 3) 4! a , (13) and for n = 3 we have a 5 = − ( p − 3)( p + 4) 5 · 4 a 2 = ( p − 1)( p − 3)( p + 2)( p + 4) 5! a 1 , (14) We can express this result in the form y ( x ) = a L p ( x ) + a 1 L 1 p ( x ) , (15) 2 where L p ( x ) is an even-powered series (hence defining an even function over its interval of convergence) and L 1 p ( x ) is an odd-powered series (hence defining an odd function over its interval of convergence)....
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set3 - Lecture 7 Series Solutions(cont’d Let us now...

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